groups-rings-fields

[south patrol]

as in the example above ^

[coral shale]

thinking............................

sorry firstly, this is inaccurate - defn of normal ext involves irreducibles

i need to specify irreducible for this to make sense with respect to the original Q

And then irreducible polynomials in K[x] may not also be irreducible polynomials in K1[x]

and that's where the problem arises

https://i.imgur.com/JHbAu9a.png

forgot we only consider irreducibles

At most we only have this then, ig.

[dull root]

On another note, if we have polynomials of the form x^n - a, it is easy to find all its roots, namely we take the real root a^(1/n) and take an primitive n-th root of 1, and just multiple its powers with the real root to get all the roots.

However, what if we instead modified the polynomial slightly, like
x^4 + x^2 - 2. It still has four roots, but how do we find all other other roots assuming we have a real root. For sure any complex roots must come in conjugate pairs

[coral shale]

quartic formula?

In general there is no way, as there is no formula. . .

if you have some root already, you could do long division, but in general that doesn't help much, surely?

[delicate bloom]

that one happens to be a quadratic in x^2 so it's not too bad

[dull root]

Yea, I think division is prob the easiest way to do it, even if tedious

[broken stirrup]

hi again, I don't get this problem 😐

[prisma shuttle]

can someone explain what an SPIR is

[broken stirrup]

ok i get it now

[next obsidian]

The idea is that what x does to a encodes a linear transformation

Ah okay

I was gonna say you want to think of x•- as a function on A

This is a common trick btw, representation theory is kinda built off of a very similar idea

[broken stirrup]

yeah multiplication with x will boil down to definition of linear transformation

[next obsidian]

Yup

[lavish nexus]

about grading automorphism on a Clifford algebra
does Clifford algebra has like some sort of unique factorization

[broken stirrup]

thanks chnmonkey chmoblubmia

[dull root]

If I have a field with $\sqrt(2 + \sqrt 2)$ in it, how to see $\sqrt (2 - \sqrt 2)$ is also in it. I am pretty sure it is true, but I can't find a good way to express $\sqrt (2 - \sqrt 2)$

[cloud walrus]

MasakaBakana

[next obsidian]

Uhh is this over Q?

[dull root]

Yes

[next obsidian]

Hmm so

I think you can argue like

Q(sqrt(2 + sqrt(2)) embeds into your field

And that contains the other element

Now what’s important is like

This is expressible using arithmetic over Z

So like umm… not every isomorphism necessarily maps an element k to itself

But let me explain for a second

[dull root]

Well the issue I am having is that Q(sqrt(2 + sqrt(2)) contains sqrt(2 - sqrt(2)

[next obsidian]

Oh

You have to figure that out

Lol

Idk if it does

Oh it does?

[dull root]

I think so, i dunno tho

[next obsidian]

Well

If it does

The reason it suffices is that any map sends Z to Z

As in, any isomorpjism takes 1 to 1

And then by additivity n to n

[delicate bloom]

you can work backwards from x=sqrt(2+sqrt(2)) and make the polynomial and see that both are roots

[next obsidian]

So suppose that Q(sqrt(2 + sqrt(2)) embeds into your field F via f

[dull root]

But the other root is -sqrt(2+sqrt(2)), not sqrt(2-sqrt(2)) right?

[next obsidian]

Suppose the former also has sqrt(2 - sqrt(2))

Then if x = f(sqrt(2 - sqrt(2))you see that (2 - x^2)^2 = 2

Because if you let y = sqrt(2 - sqrt(2)) then f((2 - y^2)^2) = f(2) = 2

But also that’s equal to (2 - x^2)^2

[delicate bloom]

there are 4 roots cause it's a quartic

and it's irreducible by eisenstein

[next obsidian]

Wouldn’t that require that you know that the polynomial splits in Q(sqrt(2 + sqrt(2)) tho?

[dull root]

I see that x^4 - 4x^2 + 2 is a polynomial it as a root

But then we need that the extension is a normal extension to guarantee it splits as roots. This is pretty much equivalent to showing sqrt(2 - sqrt(2) is in the extension, so it doesnt' seem any easier

[delicate bloom]

yeah true

[dull root]

I feel there is some stupid reason using just basic manipulations that it is in the field, but I can't see anything ):

[delicate bloom]

ok I think I found it, $$\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} = \sqrt{2-\sqrt{2}}$$

[cloud walrus]

Merosity

[dull root]

You brilliant genius

[delicate bloom]

lmao

[next obsidian]

I feel like

[dull root]

I feel so fucking stupid

[next obsidian]

you should be able to get this from like

writing the roots using the quadratic formula twice

[delicate bloom]

I broke the rules and just multiplied by the conjugate and once I got it I just fixed itup

[next obsidian]

I feel like that's supposed to tell you the extension is normal

[dull root]

Yea, it is clear this is a difference of squares

from the equation above

[tender mist]

Q[sqrt(a+sqrt(b))] contains sqrt(a-sqrt(b)) iff it contains sqrt(a^2-b)

[next obsidian]

do you just know that off the top of your head?

lmfao

maybe you've seen this in an algebraic number theory class or somethign?

[tender mist]

Doing that thing in general

Lol, no, I was trying to deduce some things

[next obsidian]

Gotcha

If you just knew this fact off the top of your head that would've been really funny

[tender mist]

Ahah, I hope to remember it now, so it may be funny in the future

[next obsidian]

[delicate bloom]

yeah, everyone will die laughing at the hilarity when you whip it out lol

[next obsidian]

I will die laughing

and that's all that matters

[delicate bloom]

I don't think I will remember it specifically but I'll probably remember how to derive it

[next obsidian]

actually

tbh

to derive that do you even need Q?

I feel like all you need is maybe like

char not 2

or something

maybe

maybe to be safe say characteristic not 0

anyway idk, I don't wanna think about it lol

[delicate bloom]

yeah like ultimately the trick is to get the negative sign, we need to invert it, and then clean it up

and our ability to clean it up depends on what we have lying around idk

$$\frac{1}{\sqrt{a+\sqrt{b}}} = \frac{1}{\sqrt{a+\sqrt{b}}}\frac{\sqrt{a-\sqrt{b}}}{\sqrt{a-\sqrt{b}}} = \frac{\sqrt{a-\sqrt{b}}}{\sqrt{a^2-b}}$$

[cloud walrus]

Merosity

[delicate bloom]

I'm imagining a similar trick would work for the analogous type of problem for $\bQ(\sqrt{a+\sqrt{b+\sqrt{c}}})$

[cloud walrus]

Merosity

[shell brook]

is this missing words?

[coral shale]

what do you think it should say 🤔

Bit of an odd question I do think, though

[delicate bloom]

I'm reading it as saying that not having a monic quadratic factor implies it has no quadratic factor

[subtle ivy]

i mean if Z_p is its usual meaning the question is trivial i think hehe

[shell brook]

lol yeah that's what I was thinking

I'm not sure what it should say

but it's like "If not x then not x" lol

[subtle ivy]

might mean a and b are in fields which arent Z_p....

but i mean idk that's just as easy

[shell brook]

¯_(ツ)_/¯

I'll just skip it

its not a pset

[subtle ivy]

was it from a text?

[shell brook]

yeah, it's from 9th ed Gallian

It's the same question in 8th ed too

so idk

[subtle ivy]

huh... weird!

[next obsidian]

It's what Mero said right?

They're saying if you don't have a monic quadratic factor you have no quadratic factor at all

[shell brook]

Yeah okay upon squinting that is the correct reading

idk why Gallian is so resistant to just introducing the term monic lmao

[next obsidian]

Cringe

[shell brook]

I don't think he introduces it until like 3 chapters later

[next obsidian]

words r scary

😭

[delicate bloom]

I think I read that question 8 times confused how it wasn't a tautology

[next obsidian]

yeah

it wasn't until you said something I realized what it was actually asking

[delicate bloom]

the question just seems overly specific too

like might as well just do it for any field and any degree polynomial

it's equally as hard lol

[devout crow]

so, you can't put a ring structure on the group Q/Z because the identity would have finite order which leads to a contradiction

but this argument doesn't work for R/Z

so can we turn this into a ring? or is there a different reason why not

[coral shale]

so R/Z is real numbers mod 1

you could say S1, the circle

[devout crow]

yeah

[coral shale]

whys this not a ring already?

oh, we have 1 problems

[devout crow]

you have the "ring addition" given by addition mod 1, or equivalently by multiplication in S1

the problem is in giving it a multiplication that distributes, etc

or even showing it exists

[coral shale]

forgetting about 1, does everything else work out 🤔 ???

oh nvm

[woven delta]

So you are being vague about something

Namely what is the particular Z you are modding out by

[devout crow]

wdym

[woven delta]

I think it kinda does matter

[devout crow]

we can work with the group S1 and drop the whole R/Z stuff

[coral shale]

I'm a bit confused why 'normal' multiplication doesn't work, other than no identity . . .

[woven delta]

Sure if you want

But then what do you want to be your identity?

[coral shale]

rng

If we just want a rng, we are good, right?

[woven delta]

What is your multiplication

But yeah I guess you are figuring that out

[devout crow]

i don't have any multiplications, or multiplication candidates

i don't even know if S1 admits one

[coral shale]

[x][y] = [xy]

[.] denoting fractional part?

[devout crow]

that's kinda the question

[coral shale]

Is this not well defined

oh

[woven delta]

I know that R/Q for a particular copy of Q inside of R is isomorphic to R as a group

[delicate orchid]

yeah that doesn't seem well defined to me

[coral shale]

2 . 0.5 = 0
3 . 0.5 = 0.5

[next obsidian]

Yeah so

Z isn’t an ideal

So you’re screwed

[woven delta]

If they're talking about putting ring structures on

[next obsidian]

Regardless of issues with 1, you need closure under arbitrary multiplication to get a well defined operation

[woven delta]

Then they are screwed for a canonical one

Idk if they are screwed in general

[devout crow]

i don't mean R/Z as a ring quotient, i mean giving the quotient group R/Z a ring structure

[woven delta]

Probably they are and it's not hard to prove

[dull root]

I have a simple question. If I want to check if an finite extension Q(a,b) is a normal extension of Q, I only need to look at the minimal polynomials over Q of the generators a and b right, and see that they split as root factors?

[next obsidian]

I mean this is uncountable

[coral shale]

if we abandon normal addition, we can create a bijection between R/Z and R and we are done

[next obsidian]

Actully bijective to R

Just import that

[woven delta]

That's bad chmonkey

[next obsidian]

Why

[woven delta]

Because they want it to be compatible with the current group structure

On S^1

I think this is a classical topic

[next obsidian]

Well in that case I’m not sure

[coral shale]

Let R, S represent the sets of roots of the minimal polynomials of a and b.
Then Q(R, S) is a splitting extension for the product of the minimal polynomials (ie. normal).
If you can show Q(a, b) = Q(R, S), then you are done. (it suffices to show Q(a, b) contains Q(R, S))

[next obsidian]

I know Q/Z doesn’t admit a ring structure

Which agrees with its additive one

[devout crow]

yeah, that example is what got me considering R/Z

[next obsidian]

Wait I think this does it for you

Or err

Idk

[devout crow]

but the same argument doesn't work

[woven delta]

Is Q/Z a subring?

[next obsidian]

I was thinking you can intersect with Q/Z < R/Z

[woven delta]

It's a subgroup

[devout crow]

because it relies on every element having finite order

[next obsidian]

And give it a ring structure

[woven delta]

It may not be a subring

[next obsidian]

But there’s no reason it’s multiplicatively closed

[woven delta]

Yes

[next obsidian]

Wait yes there is

I think its because of linearity

Like

Often Z-linearity

Combined with inverses

Fixes stuff for Q

Well no nvm

1 isn’t actually 1

It’s some random element of R/Z

So yeah idk

[woven delta]

Okay I think I can make this work

You don't necessarily need 1

Just some fixed element

Fix some nonzero element r in Q/Z, and then you can write every other element of Q/Z in terms of r and Q

[next obsidian]

Why?

You can’t use multiplication

Because then you’re using the normal multiplication on Q

[woven delta]

You can use multiplication by scalars in Z

[next obsidian]

Not the induced one from R/Z

There’s no reason that works the same way

[woven delta]

Okay one sec

[coral shale]

Do we require commutativity? If not, I have an idea --- we can treat this like a module right?

[devout crow]

we don't

like a module how

[coral shale]

Think of scalar multiplication as
a . b

Induce a bijection from R/Z to R

f : R/Z -> R

a . b := f(a) . b

[woven delta]

Okay the copy of Q/Z should satisfy the property that for your fixed nonzero r for every nonzero element q, there are numbers n and m so that nr=mq

[next obsidian]

Don’t think so

[woven delta]

Hmm

[next obsidian]

I think you’re inherently building on the idea that n•x = x + … + x

n-times

[devout crow]

i'm a little lost why are we considering Q/Z

[next obsidian]

That isn’t true

[woven delta]

Yes

Okay cool

[next obsidian]

Because 1 is not the multiplicative identity

In your fucked up R/Z ring structure

[woven delta]

Okay yes rational rotations

Smh

[coral shale]

ah this probably fails right identity. and we need bijection to be nice

[woven delta]

Sorry I do this sometimes chmonkey

[next obsidian]

It’s alright

This is what I was saying at first about

Z-linearity then inverses

[woven delta]

Right

[next obsidian]

Then I realized 1 isn’t actually the 1

In fact as an element 1 = 0

So it can’t be the identity

(Well, the multiplicative one at least )

[devout crow]

we know it can't be continuous

which is less nice

[next obsidian]

Hey question

These numbers are in bijection with S^1

Like as a group

Embedded in C

[woven delta]

Yes

[next obsidian]

And the operation is multiplication in C

[devout crow]

yes

[next obsidian]

If you have two things on the unit circle

a and b

Is a^b also modulus 1?

[devout crow]

true, but no identity

[next obsidian]

Wut

[woven delta]

How are you defining a^b?

[next obsidian]

Complex numbers

Or wait is there branch issues?

[devout crow]

as in, the left and right identities aren't the same right

[next obsidian]

Uhhh

Fuck

Sadge

[woven delta]

Yeah that's why I asked

[coral shale]

$e^{ia}\bs+e^{ib}:=e^{i(a+b)}$

[coral shale]

So one way to think about it is we have this right

[next obsidian]

I was thinking like “addition” was actually multiplication

And now we need “multiplication”

And I thought that exponentiatioj might work cuz I think it factors right

[coral shale]

yh i see where ur going

[woven delta]

How about that on the R/Z model

[next obsidian]

Okay but

What if we say

1 doesn’t need to exist

🤔

Does it then work?

[devout crow]

that seems more possible

i think that works then

[woven delta]

What works

[devout crow]

wait it's not even associative is it

[coral shale]

$z^{\cos\theta+i\sin\theta}=z^{\cos\theta}z^{i\sin\theta}$

[next obsidian]

Fugg

[next obsidian]

Gah!

[tribal moss]

What are we trying to do?

[woven delta]

Maybe you should do this logically

[coral shale]

This? I think modulus issues right

[next obsidian]

Take R/Z as an additive group

[woven delta]

Like choose elements that will work

[next obsidian]

Give it a ring structure

Or even a rng structure

Wait

Oh nvm

[tribal moss]

Rng is easy: every product is 0.

[next obsidian]

Yeah but

So booooring

[coral shale]

yh but no one said it

[next obsidian]

I assumed we wanted one thAt isn’t that lol

[coral shale]

i didnt even know this was possible

[woven delta]

It may not be

[devout crow]

i read an article and now i'm convinced rings morally need identities

so ring structure is preferable

[woven delta]

Oh I see

Okay so let's say we try to build a multiplication

[next obsidian]

Fuck that

I have become convinced that

[woven delta]

Lol that's the sort of shit I do chmonkey

[next obsidian]

If one existed it has a natural way to be seen

[woven delta]

That's silly

There's a ring structure on R/Q

[next obsidian]

Via considering these the modulus 1 elements of C

[woven delta]

Which is nontrivial

[next obsidian]

But that isn’t S^1

Idk

Lol

[woven delta]

And cannot be seen

[coral shale]

thats fine tho, just take angle

right

[woven delta]

Because R/Q is isomorphic to R as a group, but not naturally

[next obsidian]

I mean I think it should appear some way inside of C

And its arithmetic idk

[coral shale]

the addition we have rn, is just complex multiplication

[next obsidian]

Maybe there is one

But trying to manually cook it up?

[woven delta]

Lol

[next obsidian]

Like what do you even do?

[woven delta]

You build it

[next obsidian]

Hurb

[woven delta]

[next obsidian]

I’m team doesn’t exist tbh

[woven delta]

I think either there's a nice proof it doesn't exist or there's a gross construction of one

That's my opinion

[devout crow]

i'm leaning this way too

[tribal moss]

So, as an abelian group, I believe R/Z is Q/Z direct sum uncountably many copies of Q, right?

[woven delta]

That doesn't induce the ring structure

[next obsidian]

Uhhhhhh

[woven delta]

On Q/Z

[next obsidian]

Is it?

Lmfao

[devout crow]

why is this

[next obsidian]

I think it’s like

[tribal moss]

No, just trying to wrap my head round the group algebraically first.

[next obsidian]

Take the obvious Q/Z

[woven delta]

R is uncountably many copies of Q

And mod out one of them by Z

[next obsidian]

Everything else is an uncountable Q-vector space

Cuz irrational

And can’t ever pop back into Z

Or some shit

[devout crow]

why don't you mod them all out by Z

idk why this is not clicking

[woven delta]

Well you can't do all of them

You could do finitely many

If you wanted

[next obsidian]

Actually I’m not convinced this is true anymore

[woven delta]

Maybe countably many?

[tribal moss]

If alpha is irrational then alpha·Q maps injectively into R/Z.

[woven delta]

Finitely many

[next obsidian]

I just don’t think it has that nice direct sum decomposition

[woven delta]

What do you mean

[devout crow]

ok sure i can see this

[next obsidian]

Like the idea I think is look at Q/Z inside of it

Then look at all the irrational bits

But like the irrational bits aren’t closed under addition

Like 3 + pi - pi = 3

Idk

[tribal moss]

If we take R/Z and quotient out Q/Z, then I think we ought to get something isomorphic-but-not-naturally-so to R/Q.

[next obsidian]

I think this is true

But I don’t know it splits

[woven delta]

Yes

I thought of this earlier

And that R/Q is R

[next obsidian]

That’s really where my hangup is

[woven delta]

It shouldn't split

[devout crow]

a quotient A --> B --> C splits if B = A \oplus C, right?

[next obsidian]

Yes

I mean that shouldn’t be your definition

[woven delta]

Yeah that's a rough definition

[next obsidian]

Also those maps need to then be “the obvious ones”

[devout crow]

this is wikipedia recall

what's the def

[woven delta]

Because it doesn't tell you what A and C are inside of B

[next obsidian]

When you replave B with A (+) C

A splitting is a one-sided inverse

You want a map from R/Q -> R/Z such that composing this with the map R/Z -> R/Q is the identity on R/Q

Hmmm

[woven delta]

I think you mixed up the maps

[next obsidian]

Okay wait

No?

Our diagram is

0 -> Q/Z -> R/Z -> R/Q -> 0 yeah?

We want the last map to split

[woven delta]

No, I don't think so

One sec

[tribal moss]

Do we know that R/Q is isomorphic to R?

[next obsidian]

Emma says it is

[woven delta]

Yes

Because R is an uncountable product of Q's

[next obsidian]

Okay I am really really confused now cuz

[woven delta]

We are assuming that we are choosing a nice copy of Q though

[next obsidian]

Ah nvm

[woven delta]

Inside of R

[next obsidian]

I mean I assume we are picking the obvious one

[woven delta]

Sure

[tribal moss]

Right, R is a continuum-dimensional Q-vector space, so that structure is still there when we consider only addition.

[woven delta]

Yes

0 -> Q/Z -> R/Z -> R/Q -> 0 yeah?

This is what chmonkey wrote

[next obsidian]

Yes

[tribal moss]

So if we choose 1 as one of the basis vectors, when we quotient out Z it only affects that coordinate (where all the integers live).

[next obsidian]

If this splits what Tropo wrote is correct

But I think there’s no way this splits personally

[woven delta]

Okay sure sorry I was being silly chmonkey

Your exact sequence is fine

[next obsidian]

Okay I think I can probably prove it doesn't split

[tribal moss]

Not even if you consider the quotienting done in the Q^continuum view of the group?

[next obsidian]

but I need a second

¯_(ツ)_/¯

oh well

the issue is these aren't Q-vector spaces

R/Z is not a Q-vector space

because Z isn't a Q-subspace

[woven delta]

Yeah splitting for Q vector spaces vs splitting for abelian groups is very different

[tribal moss]

No, but we can still represent R as the direct group sum of continuum many copies of Q.

[next obsidian]

I mean splitting for vector spaces is free

uhhh

oh hmm

[woven delta]

I agree with that tropo

[next obsidian]

oh huh sure

[woven delta]

But the other two aren't

[next obsidian]

But hmm

[tribal moss]

And when we then quotient out a subgroup of just one of the summands, it should only affect that summand.

[next obsidian]

right

[woven delta]

Q vector spaces

Are they?

[next obsidian]

but I don't think if you do it that way

that this is isomorphic to R/Z

in the standard way anymore

is my guess

[woven delta]

That's fine

[next obsidian]

no like

it isn't

[tribal moss]

I'm not even considering that anymore.

[next obsidian]

because it won't be the circle group

this whole thing was to get a ring structure on the circle group

[woven delta]

Yes

[next obsidian]

if you put a different group structure on R/Z

[woven delta]

Okay yeah the other 2 aren't Q vector spaces

[next obsidian]

then you haven't done what you wanted

[tribal moss]

The group structure is the same.

[next obsidian]

I mean Tropo is saying

R = Q^omega

look at Z being Z + 0 + 0 + ...

[woven delta]

Q^c

[next obsidian]

wait no

so just Z in the first coordinate

then the quotient of R/Z = Q/Z + Q + Q + ...

but is that still isomorphic to the circle group??

[next obsidian]

oh yeah

lol

[woven delta]

Whatever

[next obsidian]

OH

I SEE

Tropo you're so smart

lmfao

[tribal moss]

I'm hoping this view could lead us to a conclusion about whether an interesting ring structure could possibly exist from an algebraic point of view. If we get a positive answer to that, we can try to see if we can make it look not-wild back in the usual circle group representation.

[next obsidian]

"take 1 as a basis element"

you can choose how R breaks up as Q^c

via picking a basis

[woven delta]

Yes

[next obsidian]

just pick the smart basis which has 1 as one of the basis elements

[woven delta]

Pick a basis that extends 1

Lol

[next obsidian]

Sorry, I was being silly

[woven delta]

[tribal moss]

🤔

[next obsidian]

Well now the question is can you put a ring structure on Q/Z + Q^c

which is really just Q/Z + R

[woven delta]

Yes

[devout crow]

you can't put one on Q/Z

[next obsidian]

yeah so you can't

[woven delta]

But that doesn't answer the question

[next obsidian]

but I'm not sure if a ring structure on a direct sum of groups

[woven delta]

Because you can have mixing

[next obsidian]

decomposes

yeah

[woven delta]

Because of closure under multiplication

[next obsidian]

WAIT

HOLY SHIT

I know how to do this

holy fuck

[woven delta]

[next obsidian]

I think

oh no

AAAAAAA

[woven delta]

Lol

[next obsidian]

I need Q/Z to be an R-module

noooooooooooooo

[woven delta]

Haha

Good luck with that

Lots of finite order elements

[next obsidian]

damn it

[tribal moss]

It's too small, too.

[woven delta]

I think you can say something a little nice

[next obsidian]

in general you can turn R + M into a ring for M an R-module

you let

(r,m)(r',m') = (rr',rm' + r'm)

[woven delta]

Lol

[next obsidian]

I mean

[tribal moss]

That looks vaguely like a universal solution, even.

[next obsidian]

it would even suffice if Q/Z was a Q-module

but it isn't

wait

no

fugg

do you know if Q has any decompositon

as Z + T

for any group T

[tribal moss]

I don't think it can.

[next obsidian]

rip

[delicate bloom]

we trying to make Q/Z into a ring?

[tribal moss]

There wouldn't be anyhting that's half of 1+0.

[next obsidian]

no

[woven delta]

R/Z

[next obsidian]

we know we can't

We're trying to turn R/Z into a ring

and we've decomposed it as Q/Z + Q^c

as a group

[delicate bloom]

try to turn Q_p/Z into a ring for me while you're up, ping me when you're done, goodnight

[woven delta]

Based mero

Just throws work at us

[next obsidian]

Walks in

Do this

Refuses to elaborate

Leaves

https://tenor.com/view/gigachad-chad-gif-20773266

[delicate bloom]

I mean how hard could it be, they're both just completions of Q so same proof should work for all of them

[tribal moss]

Q_p/Z_p is a subgroup of Q/Z, so it's not completely out of a hat.

[delicate bloom]

[tribal moss]

But now Q_p/Z ?

That feels weird.

[woven delta]

Okay let's go back to the Q/Z inside of R/Z thing

Well I'm going to, I'm actually going to think about this for a couple of minutes

Before I say anything

[tribal moss]

I suppose the first order of business is to decide what our unit in Q/Z oplus R should be.
Wlog it is either (1/n,0) or (1/n,1) or (0,1).
The first of these is out because it would give our ring a characteristic that its additive structure doesn't agree with.
The last feels most promising, but I can't seem to immediately rule out the middle one.

[woven delta]

I actually think that analyzing how the multiplication has to act on Q/Z is a good idea

Like what conditions does (1/p)^n have to satisfy concerning 1/p

(where ^n is iterating the multiplication we define)

(for a prime p)

I think there's actually a fairly limited set of conditions that we can specify

[devout crow]

christ almighty this discussion has gone on for over an hour

i hope to see an answer when i wake up tomorrow

[woven delta]

But I guess we aren't approaching the question with the intent to do that sort of work lol

[devout crow]

i'm tempted to email my professor about it

[tribal moss]

Call it recreational math.

[woven delta]

Yeah this is definitely recreational math

Yeah actually I think I have an answer

And it's pretty stupid

[tribal moss]

Any product of an element of Q/Z with anything must be a torsion element, meaning it's purely in the Q/Z summand, so Q/Z would have to be an ideal in the ring.

[woven delta]

Yes

Q/Z is all the torsion elements

[next obsidian]

wait what?

[woven delta]

So we are done

Okay wow

This is dumb

[next obsidian]

okay sure

why are you done?

[woven delta]

So Q/Z is closed under multiplication

[next obsidian]

uhhh

[tribal moss]

It's also closed under addition.

[woven delta]

Because any product of it is torsion

[next obsidian]

wait

I don't buy this

[woven delta]

And it is all of the torsion

[next obsidian]

give me amoment

so your idea for the product of something in Q/Z with anything being torsion

let r be in Q/Z, then r + r + ... + r n times is 0

sure

[devout crow]

any ring multiplication?

[woven delta]

Yes

That's what we just showed

[next obsidian]

so if d is any element

[woven delta]

Because of torsion

[tribal moss]

Any multiplication that distributes over addition.

[next obsidian]

god damn

yeah

[woven delta]

Lol this is what I was thinking of tropo

[next obsidian]

I thought secretly it might rely on

[woven delta]

But you beat me to it

[next obsidian]

nx = x + ... + x n-times

but it doesn't

[woven delta]

With the torsion thing

[tribal moss]

However, that in itself just makes Q/Z a rng, not necessarily a ring.

[next obsidian]

true lmao

I don't know that's a contradiction

[tribal moss]

It could have the everything-is-zero multiplication viewed by itself.

[next obsidian]

right

hurb

[woven delta]

Yeah

That's fair

Oof

Okay let's define multiplication on the rest by defining it modulo Q/Z

Does that actually give us a consistent notion

[next obsidian]

uhhh

well I mean

[woven delta]

If we know that restricted to Q/Z the multiplication is 0

[next obsidian]

it's Q^c

[woven delta]

Yes

[next obsidian]

wait this was so dumb

okay so

[woven delta]

Yes it was

[next obsidian]

like

tuples

[woven delta]

Holy shit

[next obsidian]

yeah?

(a,b)

a is in Q/Z

b is in Q^c

[woven delta]

Yeah just do R on Q^c

And 0 on Q/Z

[next obsidian]

just do (a,b)(c,d) = (0,bd)

yeah

[woven delta]

Or something

[tribal moss]

So, hm, can we build a quotient of a huge polynomial ring with a ideal that behaves like Q/Z and quotients down to something like R/Q or Q?

[next obsidian]

So the identity is

just (0,1)

right?

wait no

wait a second...

[tribal moss]

Nope, not an identity

[next obsidian]

if you wan an identity

[woven delta]

Sorry does this actually split

[next obsidian]

uhhhhhhh

Idk anymore man

why am I thinking about this

[woven delta]

I said yes before but now idk

Lol

Yeah same

Doing math on the spot is just delusional

You rapidly go between states of being sure and knowing you were wrong

Or at least I do

Okay yeah it does split

Smh

Theres no multiplicative identity

And every rng structure on R/Z is just a choice of ring structure on R that's compatible with standard addition

So that's the boring answer

Nothing to see here really

[next obsidian]

I'm not convinced there can't be a multiplicative identity

[woven delta]

I am

[next obsidian]

You can have some weird crossing stuff

If the operations stay within each coordinate, sure

[woven delta]

Oh I see

Okay let me try to show why that can't happen

I guess some interaction between Q/Z and R would be neat

[uncut girder]

Q/Z

[woven delta]

That respected the fact that Q/Z has the 0 multiplication

I think we have linearity though

[tribal moss]

I'm not sure we have concluded yet that the multiplication internally in Q/Z must be trivial, have we?

[woven delta]

In the addition component

We have

[tribal moss]

Oh well.

[next obsidian]

Uhh

I don't think we have

we know it doesn't have a 1

but it doesn't rule out some fucked up rng structure

[woven delta]

Wasn't the original thing that there's no rng structure on Q/Z

Oops if not

[next obsidian]

no

Ring

[woven delta]

Okay let me give the proof then

So I can show that the only nonzero ring structure you can put on it is induced by the standard multiplication

I think

[next obsidian]

ring or rng?

[woven delta]

Multiplication structure

[next obsidian]

We know there’s no ring structure

Yeah but like without identity?

[woven delta]

Yes

[next obsidian]

Okay

[tribal moss]

"Induced by the standard multiplication" doesn't immediately make much sense for Q/Z since Z is not an ideal of Q.

[woven delta]

Yes

Which is why it doesn't work

Okay sorry give me a minute

To write something coherent

Instead of something vague

Okay so we take the elements p/q and p'/q'. Assume p, p'<q, q'. (I'm going to assume commutative multiplication). Then we can say that their product must have that if you multiply by qq' you get 0, so a factor of qq' must be in the denominator (writing the product as an element of Q/Z). Now looking at the numerators, we can say that (pp')(1/q)(1/q')=(p/q)(p'/q'), so we know that the old numerator is a factor in the product, while the new denominator is a factor of the product of denominators

[next obsidian]

I don't believe that if you multiply by qq' it must be 0

[woven delta]

Yes sorry

It must be a natural number

[next obsidian]

eh?

I still don't see why that's true

[woven delta]

Because of torsion

[next obsidian]

so you're saying replace qq' with "a natural"

[tribal moss]

I think I have it:
If our ring has a unit, then it also has an element h such that h+h=1 (namely, just halve each component of whatever the unity is).
Now (½,0)·h + (½,0)·h = (½+½,0)·h = (0,0)·h = 0.
On the other hand (½,0)·h + (½,0)·h = (½,0)·(h+h) = (½,0)·1 = (½,0)
But 0 != (½,0).

[next obsidian]

not that multiplying by qq' results in a natural?

[woven delta]

No I am saying that multiplying by qq' is a natural

[next obsidian]

Yeah I don't buy that

Like the only way I see that you could conclude that is somethin glike

[woven delta]

Because if you give q to the first one and q' to the second one you get 0's in Q/Z

[next obsidian]

but I don't buy that

why is p/q x q = 0

[woven delta]

Because addition still works as it's supposed to

[next obsidian]

this becomes

[woven delta]

And we are living in Q/Z

[next obsidian]

p/q x 1 added q times

[tribal moss]

This argument doesn't even need the Q/Z + R decomposition, we can do it directly in R/Z.

[next obsidian]

but p/q x 1 isn't p/q

what if p/q x 1 is like p/13q or some shit

[woven delta]

Chmonkey we are living in Q/Z

So 1 is 0

[next obsidian]

[woven delta]

Lol

[next obsidian]

okay right

but wait like

isn't multiplying by qq' just multiplying by 0?

[woven delta]

No

We are adding

Oh wait

No this is fine

Saying "add this many copies together" is not the same as multiplying by 0

[next obsidian]

nice

[woven delta]

Okay that's better

[next obsidian]

okay but you wrote "multiply by qq'" so I thought that's with the induced multiplication

like

[woven delta]

Yes sorry

[next obsidian]

add n times

yeah idk it's hard cuz you have to be specific lmao

[woven delta]

Yeah I'm not that great at that

When I'm doing problems

[next obsidian]

okay I ee

also I need to say I officially do not care anymore cuz I've had to do crap for the past > hour

[woven delta]

Lol

[next obsidian]

I don't even like noncommutative, non-unital rings

[woven delta]

Okay I guess what me and tropo were doing is actually different

I was trying to show there can't be any interaction between the two things at all

I have things to do too lol

And am also procrastinating

But okay let me finish up what I'm trying to argue

Oh no I was trying to prove that there can't be any nontrivial rng structure on Q/Z

Lmao

[tribal moss]

Direct argument without decomposing:
For x in R, let [x] mean the coset x+Z in R/Z. Suppose # is a multiplication on R/Z that makes it into a (not necessarily commutative) ring with unit [u].
Then [0] = [0]#[u/2] = ([1/2]+[1/2])#[u/2] = [1/2]#[u/2] + [1/2]#[u/2] = [1/2]#([u/2]+[u/2]) = [1/2]#[u] = [1/2] which is absurd.

[next obsidian]

I couldn't resist chmonkaS

[woven delta]

Okay so I showed that (p/q)(p'/q') = (pp') (1/q)(1/q')

[next obsidian]

so okay, there's no ring with unity strcuture

so you could just take any rng structure on Q/Z (maybe there's only the trivial one, maybe not idc)

[woven delta]

Now what happens if I add a factor to both p and q

[next obsidian]

then just do the multiplication component-wise

[woven delta]

Yes

[next obsidian]

writing it as Q/Z + Q^c

cool

[woven delta]

Yes

We got that much at least

[next obsidian]

Swag

[woven delta]

Okay so if I add a factor r to both p and q I can write (p/q)(p'/q)=(pp'r)(p/qr)(p'/q')

I think

Where the pp'r multiplication is adding pp'r times

[uncut girder]

Viewing Q/Z as a Z module

[woven delta]

r(1/pr)=1/p

Yeah this is the whole thing

Smh

Yes

[tribal moss]

Hmm, first of all I think "decomposable tensor" are the last words in Defn 5.10 and the subsequent ramble is just discussion.

[woven delta]

A model of the tensor product means it satisfies the universal property of tensor products I think

[lavish nexus]

take the map phi: SU(2)XSU(2) -> GL(4,R)
sending (A,B) to the map (v -> A*vB) where v is in S^3 (treated as SU(2))
is there a clean way to show the determinant of phi(A,B) is positive? Hence in SO(4)

(not by calculating how it acts on the basis it's too painful)

[tribal moss]

Yeah, I can't get much of a point from it either. It just seems to be a roundabout way of saying you can pick a different space of the right dimension to use as the tensor product, but that was already said shorter as "up to isomorphism".

[woven delta]

They are unpacking the definition of the unique factorization property maybe

Oh this is actually useful

In terms of understanding things like Kronecker products maybe

I feel like they're using a different definition of the tensor product than is usually used

It's equivalent though

But it's phrased slightly differently

Instead of saying that for any multilinear map from the product there exists a unique linear map from the tensor product they say for any linear map from the tensor product there exists a unique multilinear map

Just thought I'd point that out

[tribal moss]

Hmm, that doesn't sound right -- it doesn't even characterize the tensor product but just says composition of two maps gives a unique result.

[vestal snow]

Since tensoring by N is an exact functor and computing homology involves finite limits and colimits only, the result follows

Is this a sufficient answer for this?

[next obsidian]

I mean idk what finite limits and colimits has to do with it but like

ultimately all you have to do is show that

[vestal snow]

exact functors commute with finite limits and colimits

[next obsidian]

oh well

sure

I mean I just think of it as like

M/N (x) K = M (x) K/ N (x) K

and you can show this really easily

like right exactness plus the fact that N (x) K actually embeds in M (x) K

let's you do it

¯_(ツ)_/¯

[vestal snow]

But then I can't use abstract nonsense unnecessarily

[next obsidian]

I mean

if you've proven that then sure

lol

[minor badger]

Hi, are non-associative algebras real algebras or just wannabe associative algebras?

[next obsidian]

it depends on who you are

if you like commutative algebra (me) then they're nothing, they're horrific little wannabe losers

but I think rep theorists and stuff consider them often enough that they specifically will say "associative algebra" when they want them to be associative

[lavish nexus]

what does $\Lambda^2 Pol_d(\mathbb{C}^2)$ look like

[cloud walrus]

Iteribus

[uncut girder]

Pol_d is degree d polynomials ?

Explain the notation

[lavish nexus]

yeah

so spanned by x_1^d_1x_2^d_2, d_1+d_2 = d

[next obsidian]

These are free C-modules on the set of monomials

and from there the exterior power has a nice basis

[uncut girder]

So when d=1, Pol_1(C^2) has basis
x, y
So /^2 Pol_1(C^2) has basis
x/\y

When d=2, Pol_2(C^2) has basis
x^2, xy, y^2
So /^2 Pol_1(C^2) has basis
x^2 /\ xy, x^2 /\ y^2, xy /\ y^2

In general, Pol_d(C^2) has dimension d+1, so /^2 Pol_d(C^2) has dimension d+1 choose 2

[lavish nexus]

right
I got the same dimension

[uncut girder]

Nice jargon

[next obsidian]

Is that not clear?

This is legitimately the cleanest and simplest way I would explain this to someone

[uncut girder]

Why say free C module when u could say vectorspace

[next obsidian]

oh

I guess you're righ tlol

[uncut girder]

Also all vector spaces are free

[next obsidian]

Honestly I said C-module because I just came from writing up a proof where I had a ring C and was dealing with free C-modules

[lavish nexus]

yes I think I see what this problem wants me to do with \Lambda^2 Pol_d(\mathbb{C}^2)

TY

[barren sierra]

So I just learned about Noetherian Rings

prof gave two definitions

1: Every ideal is finitely generated
2: Every ring has the ascending chain condition for ideals

which way tends to be the more useful way to view Noetherian Rings?

[thorn delta]

i think it just depends on the circumstance. The equivalent conditions don't end there either. At least one other useful one you can add to the list is that every set of ideals has a maximal element

[next obsidian]

both

another useful definition is "every finitely generated module is finitely presented"

like, there's a number of equivalent definitions, all are useful

[barren sierra]

I see

At least one other useful one you can add to the list is that every set of ideals has a maximal element
isn't that just ascending chain condition?

[thorn delta]

i mean, the claim is that they are all equivalent conditions. but maybe you are conflating the conclusion of Zorn's with ACC?

[barren sierra]

oh wait

yea I was

[vestal snow]

@next obsidian Are you sure can do this without using the fact that kernel and cokernel are finite limits and colimits respectively?

[next obsidian]

I mean, yeah?

[vestal snow]

I would think you'd get something like Im(d_i) (x) N which you would need to show is isomorphic to Im(d_i (x) 1)

[next obsidian]

Show that it implies this for SESes

and you're fine

I'll just write it on paper

[vestal snow]

Hmm okay

[next obsidian]

cuz I can't be fucked to tex it

[vestal snow]

This is what I have

In order to continue, I would need to pull the N in to both the numerator and denominator

And doing that requires - (x) N commuting with Im and Ker

I.E. ker(coker(-)) and ker(-)

[vestal snow]

Yeah that's what I did here too

[next obsidian]

that's it

you're now done

[vestal snow]

If you write it out though, you would get $\frac{Ker(i) \otimes N}{Im(i)\otimes N}$

[cloud walrus]

Finitely Many Bananas

[next obsidian]

what's i?

[vestal snow]

You still would need to show that $\frac{Ker(i\otimes id_N)}{Im(i\otimes id_N)}\cong \frac{Ker(i) \otimes N}{Im(i)\otimes N}$

[next obsidian]

wut

I don't see what there's left to do

[cloud walrus]

Finitely Many Bananas

[next obsidian]

I don't even know what i is or how you can possibly take that quotient

[vestal snow]

i is the inclusion of the image into the kernel

[next obsidian]

okay well

im(i) (x) N is always im(i (x) N)

and ker(i) (x) N = ker (i (x) N) when N is flat

for the first one you just exactness again

likea

all of this follows by the exact sequence I drew actually

like for the image equality

it's just the right exactness

im(f (x) N) = im f (x) N

because the map surjects

and the thing for kernel follows basically by doing the same thing, except you need flatness

It’s messy cuz I wrote over shit I already did

But you see that exactness of this identifies that first module ker f (x) K

With the kernel of f (x) K

[vestal snow]

What does this even mean?

[next obsidian]

that's like

f (x) id_K

i just wrote it that way cuz the functor is - (x) K

and it's normal to like put the maps into the functor too

like F(M) and F(f)

M an object or f a map

the last map there

N (x) K -> L (x) K

I wrote the otimes so shit on the paper

it just looks like a speck lol

[vestal snow]

Hmmm

I feel like something is being swept under the rug here

[next obsidian]

not really

If you want to make sense of what like

M/N (x) K = (M (x) K)/(N (x) K) is

this is what it turns into

idk what to say other than you just have to keep thinking abou tit

I think I remember being confused about if this actually works

[vestal snow]

I buy your argument for M/N (x) K = (M (x) K)/(N (x) K)

[next obsidian]

oh

well

that's like the exact same thing as this

[vestal snow]

I bought it from the beginning

[next obsidian]

this ting

like the = sign there

strictly doesn't make sense

but to me this says "the latter is a thing with a univ property"

and the former satisfies it

Like explicilty

in the thing I drew

we have a map from ker f (x) K -> N (x) K

[vestal snow]

I think a lot of the confusion I'm having here is with the fact that kernel is not just a module, but has a map associated to it (which is the inclusion)

[next obsidian]

this lands inside the kernel of f (x) K

yeah so

what I'm really saying is

(ker f (x) K) -> N (x) K

is a kernel of N (x) K -> L (x) K, the map f (x) K

this just follows by exactness of this diagram

Like 0 -> A -> B -> C

[vestal snow]

I just don't see how you get around the fact that you can pull - (x) N into the kernel/Image without somehow using that these are finite limits/colimits

[next obsidian]

is exact iff A -> B = ker(B -> C)

idk man

you just do this lol

being a kernel or cokernel can be tested using that diagram I wrote down

in an abelian category

or for cokernels

A -> B -> C -> 0

exact iff (B -> C) = coker(A -> B)

Like if you want a proof I can send you a document I wrote

doing homological algebra in abelian categories

[vestal snow]

Nah its np

I'm being too detail oriented

[next obsidian]

No that’s good

[vestal snow]

I understand why it is true and how to prove it

[next obsidian]

Lol

But like my argument is sound

You just have to figure out why

And I feel like it’s a “just sit on it”

[vestal snow]

Yeah I agree

[vestal snow]

I disagree with this though

I think if you wanted to write out an argument explicitly, you would be doing a diagram chase

It would probably be pretty straightforward

But I think you would essentially end up proving that exact functors commute with finite limits/colimits

Or well

[next obsidian]

I really don’t think so

[vestal snow]

You can soup up your argument to prove that

[next obsidian]

Like

[vestal snow]

Like the ideas in either proof would be the same

[next obsidian]

Idk lol my thing will furnish you isomorphisms between two exact sequences

Which say what you want it to

[vestal snow]

You'll just be working with specific examples of finite limits and colimits

[vestal snow]

Arbitrary products of modules commutes right?

$\prod_{i\in I}(\prod_{j\in J} M_{i,j})\cong\prod_{j\in J}(\prod_{i\in I}M_{i,j})$

[cloud walrus]

Finitely Many Bananas

[vestal snow]

I think they satisfy the same universal property?

Though I'm a bit skeptical because that would make a problem I'm doing trivially easy

[dull root]

Can anyone check this simple argument? Let A,B be rational.
Suppose that A has degree a over Q and B has degree b over Q such that a and b are coprime. Then

Q(A,B) = ab since

1. a divides Q(A,B) and b divides Q(A,B), so we get Q(A,B) >= ab
2. Q(A,B) <= ab since A,B had degrees a,b over Q.

and so the degree of A over B is exactly a.

[hidden haven]

Yes

[vestal snow]

Thanks

[hidden haven]

That's correct

[dull root]

I also have this question. If two Galois groups were isomorphic, then would their corresponding fields be isomorphic as well? This is just a consequence of the Galois correspondence right?

[hidden haven]

No. ℚ[√p] has galois group ℤ/2ℤ over ℚ for any prime p

[maiden ocean]

The galois correspondence tells you about how subgroups of a fixed galois group relate to subextensions of a fixed extension

[chilly ocean]

I am unable to understand how the operation in the example work here of a quasigroup

I know the definition, but I am unable to get the proper intuition

[barren sierra]

Algebra prof with the big brain advice today:

"This theorem is useful when you can use it"

[pastel cliff]

"theyre gonna live, until they die" - jhin

[heavy dagger]

Is the additive group of integers isomorphic to the additive
group of rationals? (I know they are bijective but I'm having trouble with this)

[chilly ocean]

To check Isomorphism between two groups, just check whether it is bijective (one to one correspondence) mapping, and the mapping respects the given group operations

[tribal moss]

No. (Q,+) is a division group; (Z,+) is not.

[heavy dagger]

Oh, I haven't learnt that yet... can you simplify it and give a hint for an argument for someone who's just learnt Cayley's theorem

[tribal moss]

In other words, for example x+x=a always has a solution for x in Q but not in Z.

[heavy dagger]

Ohh any mapping which preserves the group operations cannot be bijective? That's the argument isnt it?

Like if f was an isomorphism, f(x+y) = f(x) + f(y) = (f(x) - 1) + (f(y) + 1) so it is not one one?

Wait nvm, made a mistake

[delicate orchid]

no tropo's example works well

if f is an isomorphism between Z and Q then consider what f^-1(1/2) must be considering that f^-1(1/2+1/2) = 2f^-1(1/2) must be f^-1(1)
should be able to get a contradiction out of that I believe

[chilly ocean]

That works well

[heavy dagger]

My brain seems to be the only thing that isn't working well

[tranquil parcel]

Isomorphisms should 'perfectly preserve' algebraic relations like 2 * f(x) = f(2x). So if I take an isomorphism $f$ from $Q$ to $Z$, we know that in $Q$, 1/2 times 2 gives us 1. This means that whatever integer we get from plugging $\frac{1}{2}$ into $f$ should double (add to itself, using the additive structures of each group) to get one. Can we find an integer which we can add to itself to get $1$?

[cloud walrus]

Michael Harp